We begin by recalling that (REF ) holds with \(\delta :=2\varepsilon \ge 0\)  (see the proof of Proposition REF ).
Setting \(G_n:=A_n+g_n\)  for \(n\in \mathbb {N}\)  and
\(\mathcal {G}(m,n):={\left\lbrace \begin{array}{ll}G_{m-1} \circ \ldots \circ G_n, & m>n;\\\text{\rm Id}, & m=n,\end{array}\right.}\) 

we easily verify that
\(\mathcal {G}(m,n)(x)={A}(m,n)x+\sum _{j=n}^{m-1}{A}(m, j+1)g_j ( \mathcal {G}(j,n)(x))\) 
for \(m\ge n\)  and \(x\in \mathbb {R}^d\) , and therefore
\(D\mathcal {G}(m,n)(x)={A}(m,n)+\sum _{j=n}^{m-1} {A}(m,j+1) Dg_j(\mathcal {G}(j,n)(x)) D\mathcal {G}(j,n)(x).\) 
Observe that (REF ) and (REF ) imply that
\(\Vert Dg_n(x)v\Vert _{n+1} \le C(n+1)^{2\varepsilon }\Vert Dg_n(x)v\Vert \le \frac{cC}{n+1}\Vert v\Vert \le \frac{cC}{n+1}\Vert v\Vert _n\) 
and therefore
\(\begin{split}&\Vert D\mathcal {G}(m,n)(x) v\Vert _m \\&\le K\bigg (\frac{m}{n} \bigg )^a \Vert v\Vert _n+K \sum _{j=n}^{m-1} \bigg (\frac{m}{j+1} \bigg )^a\Vert  Dg_j(\mathcal {G}(j,n)(x)) D\mathcal {G}(j,n)(x)v\Vert _{j+1} \\&\le K\bigg (\frac{m}{n} \bigg )^a \Vert v\Vert _n+K \sum _{j=n}^{m-1} \bigg (\frac{m}{j+1} \bigg )^a \frac{cC}{j+1} \Vert D\mathcal {G}(j,n)(x)v\Vert _j\end{split}\) 
by (REF ),
which is equivalent to
\(\bigg (\frac{n}{m}\bigg )^a \Vert D\mathcal {G}(m,n)(x) v\Vert _m \le K\Vert v\Vert _n+K\sum _{j=n}^{m-1} \frac{cC}{j+1}\bigg (\frac{n}{j}\bigg )^a\Vert (D\mathcal {G}(j,n))(x)v\Vert _j\) 
for \(m\ge n\)  and \(x, v\in \mathbb {R}^d\) . We need the following discrete-version of Gronwall's lemma (see e.g. [1]}).
Lemma 1 
Let \(n\in \mathbb {N}\)  and \(\alpha >0\) . Suppose that \((u_m)_{m\ge n}\)  and \((z_m)_{m\ge n}\)  are two nonnegative sequences satisfying
\(u_m \le K\lbrace u_0 + \sum _{j=n}^{m-1} z_ju_j\rbrace \) 
for \(m \ge n\) .
Then
\(u_m \le Ku_0\,e^{\sum _{j=n}^{m-1}(Kz_j)}, \quad \forall m \ge n.\) 

By the above lemma, we find that
\(\bigg (\frac{n}{m}\bigg )^a\Vert  D\mathcal {G}(m,n)(x)v\Vert _m\le K\Vert v\Vert _n e^{\sum _{j=n}^{m-1}\frac{cC K}{j+1}}.\) 
Since \({\sum _{j=n}^{m-1}\frac{1}{j+1}}\le \log m-\log n\) ,
we obtain that
\(\Vert  D\mathcal {G}(m,n)(x)v\Vert _m \le K \bigg (\frac{m}{n}\bigg )^{a+cC K}\Vert v\Vert _n,\) 
for \(m\ge n\ge 1\)  and \(x, v\in \mathbb {R}^d\) .
Furthermore, we see from (REF ) that
\(&D\mathcal {G}(m,n)(x)-D\mathcal {G}(m, n)(y)\nonumber \\&=\phantom{=}\sum _{j=n}^{m-1} {A}(m,j+1)Dg_j(\mathcal {G}(j,n)(x))D\mathcal {G}(j,n)(x)\nonumber \\&\phantom{=}-\sum _{j=n}^{m-1} {A}(m,j+1)Dg_j(\mathcal {G}(j,n)(y))D\mathcal {G}(j,n)(y)\nonumber \\&=\phantom{=}\sum _{j=n}^{m-1}{A}(m,j+1)Dg_j(\mathcal {G}(j,n)(x))(D\mathcal {G}(j,n)(x)-D\mathcal {G}(j,n)(y))\nonumber \\&\phantom{=}-\sum _{j=n}^{m-1}{A}(m,j+1)(Dg_j(\mathcal {G}(j,n)(y))-Dg_j(\mathcal {G}(j,n)(x)))D\mathcal {G}(j,n)(y).\) 

In addition, (REF ), (REF ) and (REF ) imply that
\(&\bigg \Vert \sum _{j=n}^{m-1}{A}(m,j+1)Dg_j(\mathcal {G}(j,n)(x))(D\mathcal {G}(j,n)(x)-D\mathcal {G}(j,n)(y))v\bigg \Vert _m\nonumber \\&\le K\sum _{j=n}^{m-1}\bigg (\frac{m}{j+1}\bigg )^a \frac{cC}{j+1}\Vert (D\mathcal {G}(j,n)(x)-D\mathcal {G}(j,n)(y))v\Vert _j.\) 

Furthermore, by (REF ), (REF ), (REF ) and (REF ) we have that
\(&\bigg \Vert \sum _{j=n}^{m-1}{A}(m,j+1)(Dg_j(\mathcal {G}(j,n)(y))-Dg_j(\mathcal {G}(j,n)(x)))D\mathcal {G}(j,n)(y)v\bigg \Vert _m\nonumber \\&\le \sum _{j=n}^{m-1} K\bigg (\frac{m}{j+1}\bigg )^a \Vert (Dg_j(\mathcal {G}(j,n)(y))-Dg_j(\mathcal {G}(j,n)(x)))D\mathcal {G}(j,n)(y)v\Vert _{j+1}\nonumber \\&\le \sum _{j=n}^{m-1} KC\bigg (\frac{m}{j+1}\bigg )^a (j+1)^{2\varepsilon } \Vert (Dg_j(\mathcal {G}(j,n)(y))-Dg_j(\mathcal {G}(j,n)(x)))D\mathcal {G}(j,n)(y)v\Vert \nonumber \\&\le \sum _{j=n}^{m-1} \bigg (\frac{m}{j+1}\bigg )^a (j+1)^{2\varepsilon }\frac{KLC}{(j+1)^{1+2\varepsilon }}\Vert \mathcal {G}(j,n)(x)-\mathcal {G}(j,n)(y)\Vert  \cdot \Vert D\mathcal {G}(j,n)(y)v\Vert \nonumber \\&\le \sum _{j=n}^{m-1} \bigg (\frac{m}{j+1}\bigg )^a \frac{KLC}{j+1}\Vert \mathcal {G}(j,n)(x)-\mathcal {G}(j,n)(y)\Vert _j \cdot \Vert D\mathcal {G}(j,n)(y)v\Vert _j\nonumber \\&\le \sum _{j=n}^{m-1} \bigg (\frac{m}{j+1}\bigg )^a \frac{KLC}{j+1}K^2 \bigg (\frac{j}{n}\bigg )^{2a+2cCK}\Vert x-y\Vert _n \cdot \Vert v\Vert _n\nonumber \\&\le K^3LC\bigg (\frac{m}{n}\bigg )^{2a+2cCK}~\sum _{j=n}^{m-1}\frac{1}{j+1}\,\Vert x-y\Vert _n\cdot \Vert v\Vert _n\nonumber \\&\le K^3LC\bigg (\frac{m}{n}\bigg )^{2a+2cCK+1}\Vert x-y\Vert _n\cdot \Vert v\Vert _n.\) 

Hence,
from (REF ), (REF ) and (REF ), we obtain that
\(\begin{split}&\Vert (D\mathcal {G}(m,n)(x)-D\mathcal {G}(m, n)(y))v\Vert _m \\&\le K \sum _{j=n}^{m-1}\bigg (\frac{m}{j+1}\bigg )^a \frac{cC}{j+1}\Vert (D\mathcal {G}(j,n)(x)-D\mathcal {G}(j,n)(y))v\Vert _j \\&\phantom{\le }+K^3LC\bigg (\frac{m}{n}\bigg )^{2a+2cCK+1}\Vert x-y\Vert _n \cdot \Vert v\Vert _n.\end{split}\) 
This together with Lemma REF  implies that
\(\Vert (D\mathcal {G}(m,n)(x)-D\mathcal {G}(m, n)(y))v\Vert _m\le \tilde{K}\bigg (\frac{m}{n}\bigg )^{\tilde{a}}\Vert x-y\Vert _n \cdot \Vert v\Vert _n,\) 
for \(m\ge n\ge 1\)  and \(x, y, v\in \mathbb {R}^d\) , where \(\tilde{K}:= K^3LC>0\)  and \(\tilde{a}:=2a+3cCK+1>0\) .
On the other hand, observe that (REF ) implies that
\(\mathcal {G}(2^{m}, 2^{n})(x)=\mathcal {B}(m, n)x+\sum _{j=2^{n}}^{2^{m}-1} {A}(2^{m}, j+1)g_j(\mathcal {G}(j, 2^{n})(x)),\) 
for \(m\ge n\ge 0\)  and \(x\in \mathbb {R}^d\) . For \(n\in \mathbb {Z}^+\)  and \(x\in \mathbb {R}^d\) , set
\(f_n(x)&:=\mathcal {G}(2^{n+1}, 2^{n})(x)-\mathcal {B}(n+1, n)x\\&=\sum _{j=2^n}^{2^{n+1}-1}{A}(2^{n+1}, j+1)g_j(\mathcal {G}(j, 2^n)(x)).\) 

Then,
\(Df_n(x)=\sum _{j=2^n}^{2^{n+1}-1}{A}(2^{n+1}, j+1) Dg_j(\mathcal {G}(j, 2^n)(x))D\mathcal {G}(j, 2^n)(x).\) 
It is clear that \(f_n\)  for \(n\in \mathbb {Z}^+\)  satisfy (REF ) by (REF ).
In order to show that \(f_n\)  satisfy (REF ) with the sequence of norms \(\lbrace \Vert \cdot \Vert _{2^n}; n\in \mathbb {Z}^+\rbrace \) , we note that (REF ), (REF ) and (REF ) imply that
\(\begin{split}\Vert Df_n(x)v \Vert _{2^{n+1}} &\le \sum _{j=2^n}^{2^{n+1}-1}K\bigg (\frac{2^{n+1}}{j+1} \bigg )^a \frac{cC}{j+1}K \bigg (\frac{j}{2^n} \bigg )^{a+cCK}\Vert v\Vert _{2^n}\\&\le K^2cC\,2^{2a+cCK} \sum _{j=2^n}^{2^{n+1}-1}\frac{1}{j+1}\Vert v\Vert _{2^n}\le \eta \Vert v\Vert _{2^n},\end{split}\) 
for \(n\in \mathbb {Z}^+\)  and \(x,v\in \mathbb {R}^d\)  , where \(\eta :=(K^2C\,2^{2a+cCK+1})c>0\) .
This proves the first inequality of (REF ).
Note that we can make \(\eta \)  sufficiently small provided that we take \(c\)  small enough. Moreover, (REF ) gives that
\(\begin{split}\Vert (Df_n(x)-Df_n (y))v\Vert _{2^{n+1}} &=\Vert (D\mathcal {G}(2^{n+1}, 2^n)(x)-D\mathcal {G}(2^{n+1}, 2^n)(y))v\Vert _{2^{n+1}}\\& \le \tilde{K}2^{\tilde{a}}\Vert x-y\Vert _{2^n} \cdot \Vert v\Vert _{2^n},\end{split}\) 
for \(n\in \mathbb {Z}^+\)  and \(x, y, v\in \mathbb {R}^d\) .
This proves the second inequality of (REF ) and therefore (REF ) is proved.
Applying Theorem REF  to \(\mathbb {B}=(B_n)_{n\in \mathbb {Z}^+}\)  given by (REF ) with the sequence of norms \(\lbrace \Vert  \cdot \Vert _{2^n}; \ n\in \mathbb {Z}^+\rbrace \) , we conclude that there exists a sequence of \(C^1\) -diffeomorphisms \(h_n\colon \mathbb {R}^d \rightarrow \mathbb {R}^d\) , \(n\in \mathbb {Z}^+\) , such that
\(h_{n+1}\circ (B_n+f_n)=B_n\circ h_n, \quad n\in \mathbb {Z}^+.\) 
Moreover, there exist \(M, \rho >0\)  such that
\(\Vert D h_n(x)v\Vert _{2^n} \le M\Vert v\Vert _{2^n}, \quad \Vert Dh_n^{-1}(x)v\Vert _{2^n} \le M\Vert v\Vert _{2^n}, \quad \forall n\in \mathbb {Z}^+,\) 
for \(v\in \mathbb {R}^d\)  and \(x\in \mathbb {R}^d\)  satisfying \(\Vert x\Vert _{2^n}\le \rho \) .
Hence,
\(h_{n+1}\circ \mathcal {G}(2^{n+1}, 2^n)={A}(2^{n+1}, 2^n)\circ h_n, \quad n\in \mathbb {Z}^+.\) 
Letting \(k\in \mathbb {N}\)  and \(n\in \mathbb {Z}^+\)  be such that \(2^n\le k<2^{n+1}\) , we set
\(\psi _k:={A}(k, 2^n)\circ h_n\circ \mathcal {G}(2^n, k)\) 
and assert that (REF ) holds. For this purpose, we first assume that \(2^n\le k<k+1<2^{n+1}\) . Then,
\(\begin{split}\psi _{k+1}\circ (A_k+g_k)&={A}(k+1, 2^n)\circ h_n \circ \mathcal {G}(2^n, k+1) \circ (A_k+g_k) \\&=A_k\circ {A}(k, 2^n)\circ h_n \circ \mathcal {G}(2^n, k) \\&=A_k\circ \psi _k,\end{split}\) 
and thus (REF ) holds. Let us now suppose that \(k+1=2^{n+1}\) . Then, using (REF ) we have that
\(\begin{split}A_k\circ \psi _k&={A}(k+1, 2^n)\circ h_n \circ \mathcal {G}(2^n, k) \\&={A}(2^{n+1},2^n)\circ h_n \circ \mathcal {G}(2^n, k)\\&=h_{n+1} \circ \mathcal {G}(2^{n+1}, 2^n)\circ \mathcal {G}(2^n, k)\\&=h_{n+1} \circ \mathcal {G}(k+1, k) =h_{n+1} \circ (A_k+g_k)=\psi _{k+1} \circ (A_k+g_k),\end{split}\) 
where the last equality holds because \(k+1=2^{n+1}\)  implies that
\(\psi _{k+1}=\mathcal {G}(k+1, 2^{n+1})\circ h_{n+1}\circ {A}(2^{n+1}, k+1)=h_{n+1}.\) 
Consequently, (REF ) again holds. Finally, using (REF ), (REF ), (REF ) and (REF ) we conclude that
\(\begin{split}\Vert D\psi _k(x)v\Vert  &\le \Vert D\psi _k(x)v\Vert _{k}\le K2^{a+cCK}\Vert Dh_n({A}(2^n, k)x){A}(2^n, k)v\Vert _{2^n}\\&\le K^2M2^{2a+cCK} \Vert v\Vert _k \le CK^2M2^{2a+cCK}k^{2\varepsilon } \Vert v\Vert \end{split}\) 
whenever \(\Vert x\Vert \le \frac{\rho }{CK2^ak^{2\varepsilon }}\)  since it implies
\(\Vert {A}(2^n, k)x\Vert _{2^n}\le K2^a \Vert x\Vert _k \le CK2^a k^{2\varepsilon }\Vert x\Vert  \le \rho .\) 
Hence, the first estimate in (REF ) holds. Similarly, one can establish the second estimate in (REF ). The proof of this theorem is completed.
